Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$

I have a partial solution, as follows:

Note that $\frac{(b-c)(1+a^2)}{x+a^2}=\frac{(b-c)\left((x+a^2)+(1-x)\right)}{x+a^2}=(b-c)+\frac{{(b-c)}(1-x)}{x+a^2}$. Likewise, $\frac{(c-a)(1+b^2)}{x+b^2}=(c-a)+\frac{(c-a)(1-x)}{x+b^2}$ and $\frac{(a-b)(1+c^2)}{x+c^2}=(c-a)+\frac{(a-b)(1-x)}{x+c^2}$.

Now, $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=\frac{(b-c)(1-x)}{x+a^2}+\frac{(c-a)(1-x)}{x+b^2}+\frac{(a-b)(1-x)}{x+c^2}$ as $(b-c)+(c-a)+(a-b)=0$.

Hence $(1-x)\left(\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}\right)=0$ and so $x=1$ or $\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}=0$

$$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)+\frac{(x+a^2)(c-a)(1+b^2)}{x+b^2}+\frac{(x+a^2)(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x+b^2)+(x+a^2)(c-a)(1+b^2)+\frac{(x+a^2)(x+b^2)(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x+b^2)(x+c^2)+(x+c^2)(x+a^2)(c-a)(1+b^2)+(x+a^2)(x+b^2)(a-b)(1+c^2)=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x^2+x(b^2+c^2)+(bc)^4)+(x^2+x(a^2+c^2)+(ac)^4)(c-a)(1+b^2)+(x^2+x(b^2+a^2)+(ba)^4)(a-b)(1+c^2)=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x^2)+x(b-c)(1+a^2)(b^2+c^2)+(bc)^4(b-c)(1+a^2)+(x^2)(c-a)(1+b^2)+x(c-a)(1+b^2)(a^2+c^2)+(ac)^4(c-a)(1+b^2)+(x^2)(a-b)(1+c^2)+x(b^2+a^2)(a-b)(1+c^2)+(ba)^4(a-b)(1+c^2)=0$$ $$\Leftrightarrow [(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)](x^2)+x[(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)]+(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)=0$$ $$\Leftrightarrow x^2+\frac{(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}x+\frac{(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}=0$$

You can now simplify and solve the quadratic.

Starting with pipi's simplification of the mess: $$ \frac{b - c}{x + a^2} + \frac{c - a}{x + b^2} + \frac{a - b}{x + c^2} = 0 \\ (b - c) (x + b^2) (x + c^2) + (c - a) (x + a^2) (x + c^2) + (a - b) (x + a^2) (x + b^2) = 0 $$ Suprisingly, this turns out a linear equation for $x$, with solution: $$ x = a b + a c + b c $$ (Many thanks to Maxima for help with algebra)